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Miscellaneous Exercise · Q1

Q.Prove that 2cos⁡π13cos⁡9π13+cos⁡3π13+cos⁡5π13=02\cos\frac{\pi}{13}\cos\frac{9\pi}{13} + \cos\frac{3\pi}{13} + \cos\frac{5\pi}{13} = 0.

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✓ Free question

Using the product-to-sum identity, the first term simplifies to cos⁡8π13+cos⁡10π13\cos\frac{8\pi}{13} + \cos\frac{10\pi}{13}. Then, pairing with the other cosines and using cos⁡(π−θ)=−cos⁡θ\cos(\pi - \theta) = -\cos\theta shows all terms cancel, giving 00.

We need to prove:

2cos⁡π13cos⁡9π13+cos⁡3π13+cos⁡5π13=0.2\cos\frac{\pi}{13}\cos\frac{9\pi}{13} + \cos\frac{3\pi}{13} + \cos\frac{5\pi}{13} = 0.

The key is to notice that the angles π13,3π13,5π13,9π13\frac{\pi}{13}, \frac{3\pi}{13}, \frac{5\pi}{13}, \frac{9\pi}{13} are all odd multiples of π13\frac{\pi}{13}. Many of these angles are supplementary to each other when added to π\pi, which creates sign cancellations. The product 2cos⁡Acos⁡B2\cos A \cos B is a natural place to start — it begs to be converted into a sum.

  1. Apply the product-to-sum identity Recall: 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A \cos B = \cos(A+B) + \cos(A-B). Here A=π13A = \frac{\pi}{13}, B=9π13B = \frac{9\pi}{13}. So:

2cos⁡π13cos⁡9π13=cos⁡(π13+9π13)+cos⁡(π13−9π13)=cos⁡10π13+cos⁡(−8π13).2\cos\frac{\pi}{13}\cos\frac{9\pi}{13} = \cos\left(\frac{\pi}{13} + \frac{9\pi}{13}\right) + \cos\left(\frac{\pi}{13} - \frac{9\pi}{13}\right) = \cos\frac{10\pi}{13} + \cos\left(-\frac{8\pi}{13}\right).

Since cosine is even, cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta, so this becomes:

cos⁡10π13+cos⁡8π13.\cos\frac{10\pi}{13} + \cos\frac{8\pi}{13}.

  1. Rewrite the whole expression The original expression is now:

cos⁡8π13+cos⁡10π13+cos⁡3π13+cos⁡5π13.\cos\frac{8\pi}{13} + \cos\frac{10\pi}{13} + \cos\frac{3\pi}{13} + \cos\frac{5\pi}{13}.

  1. Look for supplementary angle pairs

    Notice that 8π13\frac{8\pi}{13} and 5π13\frac{5\pi}{13} add up to 13π13=π\frac{13\pi}{13} = \pi.

    Similarly, 10π13\frac{10\pi}{13} and 3π13\frac{3\pi}{13} also add up to π\pi.

    This is the crucial observation.

    For any angle θ\theta, cos⁡(π−θ)=−cos⁡θ\cos(\pi - \theta) = -\cos\theta.

    So:

    • cos⁡8π13=cos⁡(π−5π13)=−cos⁡5π13\cos\frac{8\pi}{13} = \cos\left(\pi - \frac{5\pi}{13}\right) = -\cos\frac{5\pi}{13}.
    • cos⁡10π13=cos⁡(π−3π13)=−cos⁡3π13\cos\frac{10\pi}{13} = \cos\left(\pi - \frac{3\pi}{13}\right) = -\cos\frac{3\pi}{13}.
  2. Substitute these into the sum

cos⁡8π13+cos⁡10π13+cos⁡3π13+cos⁡5π13=(−cos⁡5π13)+(−cos⁡3π13)+cos⁡3π13+cos⁡5π13.\cos\frac{8\pi}{13} + \cos\frac{10\pi}{13} + \cos\frac{3\pi}{13} + \cos\frac{5\pi}{13} = \left(-\cos\frac{5\pi}{13}\right) + \left(-\cos\frac{3\pi}{13}\right) + \cos\frac{3\pi}{13} + \cos\frac{5\pi}{13}.

  1. Cancel terms The −cos⁡5π13-\cos\frac{5\pi}{13} and +cos⁡5π13+\cos\frac{5\pi}{13} cancel. The −cos⁡3π13-\cos\frac{3\pi}{13} and +cos⁡3π13+\cos\frac{3\pi}{13} cancel. Everything sums to 00.
Watch out

A common mistake is to forget that cos⁡(π−θ)=−cos⁡θ\cos(\pi - \theta) = -\cos\theta only works when the angle is exactly π\pi minus something. Here 8π13\frac{8\pi}{13} is indeed π−5π13\pi - \frac{5\pi}{13}, so it's valid. Always check the sum equals π\pi exactly.

Tip

This trick — converting a product into a sum, then pairing terms that sum to π\pi — is a powerful pattern for proving identities with angles like kπn\frac{k\pi}{n} where nn is odd. It often turns a messy expression into a clean zero.

✓Final answer

The expression simplifies to 00, so the identity is proved.

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